Mathematics, 31.05.2021 14:20, danessaporter

A two-sample t-test for a difference in means will be conducted to investigate whether the average amount of money spent per customer at Department Store M is different from that at Department Store V. From a random sample of 35 customers at Store M, the average amount spent was $300 with standard deviation $40. From a random sample of 40 customers at Store V, the average amount spent was $290 with standard deviation $35. Assuming a null hypothesis of no difference in population means, what is the test statistic for the appropriate test to investigate whether there is a difference in population means?

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answered: uehlingt39

The test statistic is t = 1.14.

Step-by-step explanation:

Before testing the hypothesis, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

From a random sample of 35 customers at Store M, the average amount spent was $300 with standard deviation $40.

This means that $\mu_M = 300, s_M = \frac{40}{\sqrt{35}} = 6.761$

From a random sample of 40 customers at Store V, the average amount spent was $290 with standard deviation $35.

This means that $\mu_V = 290, s_V = \frac{35}{\sqrt{40}} = 5.534$

Assuming a null hypothesis of no difference in population means

At the null hypothesis, we test that there is no difference, that is, the subtraction of the means is 0. So

$H_0: \mu_M - \mu_V = 0$

At the alternate hypothesis, we test if there is difference, that is, the subtraction of the means is different of 0. So

$H_1: \mu_M - \mu_V \neq 0$

The test statistic is:

$t = \frac{X - \mu}{s}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, and s is the standard error.

0 is tested at the null hypothesis:

This means that $\mu = 0$

From the samples:

$X = \mu_M - \mu_V = 300 - 290 = 10$

$s = \sqrt{s_M^2+s_V^2} = \frac{6.761^2+5.534^2} = 8.737$

What is the test statistic for the appropriate test to investigate whether there is a difference in population means?

$t = \frac{X - \mu}{s}$

$t = \frac{10 - 0}{8.737}$

t = 1.14

The test statistic is t = 1.14.

answered: Guest

answer: since the question is talking about growth you have to use the growth formula which is inital amount (1 + growth)^time

step-by-step explanation:

the intial amount is 1200, the growth is 4.8, and the time is 7. plug those numbers into the equation, also i think you have to change 4.8 into a percent. but, i have forgotton how to do that. if this is a online class go on to the lesson.

answered: Guest

80% of the cookies

step-by-step explanation: